Force of Friction

Mar 2015
1
0
Hi! I'm having trouble with solving this question. Hopefully I can get help here
A force of 112 N acts on 35 kg box which moves at a steady speed across the floor. The force is directed 40 degrees above the ground.
How to state Newton's second law using x components and y components?
x: -Fa(cos40) -Ff=ma and y: -Fa(sin40)+Fn-Fg=0?
How do you calculate force of friction and normal force from this information?
How do you find the force the box is acting on the floor (magnitude and direction)?
I really really need help with this
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
Since the box moves at constant speed across the floor you know that acceleration = 0, so your x-direction equation becomes:

112 cos(40) = -Ff, where Ff= force of friction

Your y-direction equation tells you that the normal force is Fn = 112sin(40) + mg. So you have everything you need to determnine the value of Fn.

The total force acting on the floor is then the vector sum of the negative of Ff and Fn (i.e., the forces of the box acting on the floor are equal and opposite to the forces of the floor acting on the box).
 
Last edited:
Sep 2015
5
0
How to state Newton's second law using x components and y components?
x: -Fa(cos40) -Ff=ma and y: -Fa(sin40)+Fn-Fg=0?
Placing the x axis in the direction of motion:
Fnet,x = Fa cos40° - Ff = m a_x = 0
Fnet,y = Fa sin40° + Fn - mg = m a_y = 0

Therefore Fn should be mg - Fa sin40°, and not mg + Fa sin40° (the angle is 40° with the horizontal, which means that Fa sin40° is helping Fn)